0.957 603 280 698 573 651 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.957 603 280 698 573 651(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
0.957 603 280 698 573 651(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

2. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.957 603 280 698 573 651.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.957 603 280 698 573 651 × 2 = 1 + 0.915 206 561 397 147 302;
  • 2) 0.915 206 561 397 147 302 × 2 = 1 + 0.830 413 122 794 294 604;
  • 3) 0.830 413 122 794 294 604 × 2 = 1 + 0.660 826 245 588 589 208;
  • 4) 0.660 826 245 588 589 208 × 2 = 1 + 0.321 652 491 177 178 416;
  • 5) 0.321 652 491 177 178 416 × 2 = 0 + 0.643 304 982 354 356 832;
  • 6) 0.643 304 982 354 356 832 × 2 = 1 + 0.286 609 964 708 713 664;
  • 7) 0.286 609 964 708 713 664 × 2 = 0 + 0.573 219 929 417 427 328;
  • 8) 0.573 219 929 417 427 328 × 2 = 1 + 0.146 439 858 834 854 656;
  • 9) 0.146 439 858 834 854 656 × 2 = 0 + 0.292 879 717 669 709 312;
  • 10) 0.292 879 717 669 709 312 × 2 = 0 + 0.585 759 435 339 418 624;
  • 11) 0.585 759 435 339 418 624 × 2 = 1 + 0.171 518 870 678 837 248;
  • 12) 0.171 518 870 678 837 248 × 2 = 0 + 0.343 037 741 357 674 496;
  • 13) 0.343 037 741 357 674 496 × 2 = 0 + 0.686 075 482 715 348 992;
  • 14) 0.686 075 482 715 348 992 × 2 = 1 + 0.372 150 965 430 697 984;
  • 15) 0.372 150 965 430 697 984 × 2 = 0 + 0.744 301 930 861 395 968;
  • 16) 0.744 301 930 861 395 968 × 2 = 1 + 0.488 603 861 722 791 936;
  • 17) 0.488 603 861 722 791 936 × 2 = 0 + 0.977 207 723 445 583 872;
  • 18) 0.977 207 723 445 583 872 × 2 = 1 + 0.954 415 446 891 167 744;
  • 19) 0.954 415 446 891 167 744 × 2 = 1 + 0.908 830 893 782 335 488;
  • 20) 0.908 830 893 782 335 488 × 2 = 1 + 0.817 661 787 564 670 976;
  • 21) 0.817 661 787 564 670 976 × 2 = 1 + 0.635 323 575 129 341 952;
  • 22) 0.635 323 575 129 341 952 × 2 = 1 + 0.270 647 150 258 683 904;
  • 23) 0.270 647 150 258 683 904 × 2 = 0 + 0.541 294 300 517 367 808;
  • 24) 0.541 294 300 517 367 808 × 2 = 1 + 0.082 588 601 034 735 616;
  • 25) 0.082 588 601 034 735 616 × 2 = 0 + 0.165 177 202 069 471 232;
  • 26) 0.165 177 202 069 471 232 × 2 = 0 + 0.330 354 404 138 942 464;
  • 27) 0.330 354 404 138 942 464 × 2 = 0 + 0.660 708 808 277 884 928;
  • 28) 0.660 708 808 277 884 928 × 2 = 1 + 0.321 417 616 555 769 856;
  • 29) 0.321 417 616 555 769 856 × 2 = 0 + 0.642 835 233 111 539 712;
  • 30) 0.642 835 233 111 539 712 × 2 = 1 + 0.285 670 466 223 079 424;
  • 31) 0.285 670 466 223 079 424 × 2 = 0 + 0.571 340 932 446 158 848;
  • 32) 0.571 340 932 446 158 848 × 2 = 1 + 0.142 681 864 892 317 696;
  • 33) 0.142 681 864 892 317 696 × 2 = 0 + 0.285 363 729 784 635 392;
  • 34) 0.285 363 729 784 635 392 × 2 = 0 + 0.570 727 459 569 270 784;
  • 35) 0.570 727 459 569 270 784 × 2 = 1 + 0.141 454 919 138 541 568;
  • 36) 0.141 454 919 138 541 568 × 2 = 0 + 0.282 909 838 277 083 136;
  • 37) 0.282 909 838 277 083 136 × 2 = 0 + 0.565 819 676 554 166 272;
  • 38) 0.565 819 676 554 166 272 × 2 = 1 + 0.131 639 353 108 332 544;
  • 39) 0.131 639 353 108 332 544 × 2 = 0 + 0.263 278 706 216 665 088;
  • 40) 0.263 278 706 216 665 088 × 2 = 0 + 0.526 557 412 433 330 176;
  • 41) 0.526 557 412 433 330 176 × 2 = 1 + 0.053 114 824 866 660 352;
  • 42) 0.053 114 824 866 660 352 × 2 = 0 + 0.106 229 649 733 320 704;
  • 43) 0.106 229 649 733 320 704 × 2 = 0 + 0.212 459 299 466 641 408;
  • 44) 0.212 459 299 466 641 408 × 2 = 0 + 0.424 918 598 933 282 816;
  • 45) 0.424 918 598 933 282 816 × 2 = 0 + 0.849 837 197 866 565 632;
  • 46) 0.849 837 197 866 565 632 × 2 = 1 + 0.699 674 395 733 131 264;
  • 47) 0.699 674 395 733 131 264 × 2 = 1 + 0.399 348 791 466 262 528;
  • 48) 0.399 348 791 466 262 528 × 2 = 0 + 0.798 697 582 932 525 056;
  • 49) 0.798 697 582 932 525 056 × 2 = 1 + 0.597 395 165 865 050 112;
  • 50) 0.597 395 165 865 050 112 × 2 = 1 + 0.194 790 331 730 100 224;
  • 51) 0.194 790 331 730 100 224 × 2 = 0 + 0.389 580 663 460 200 448;
  • 52) 0.389 580 663 460 200 448 × 2 = 0 + 0.779 161 326 920 400 896;
  • 53) 0.779 161 326 920 400 896 × 2 = 1 + 0.558 322 653 840 801 792;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


4. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.957 603 280 698 573 651(10) =

0.1111 0101 0010 0101 0111 1101 0001 0101 0010 0100 1000 0110 1100 1(2)

5. Positive number before normalization:

0.957 603 280 698 573 651(10) =

0.1111 0101 0010 0101 0111 1101 0001 0101 0010 0100 1000 0110 1100 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the right, so that only one non zero digit remains to the left of it:


0.957 603 280 698 573 651(10) =

0.1111 0101 0010 0101 0111 1101 0001 0101 0010 0100 1000 0110 1100 1(2) =

0.1111 0101 0010 0101 0111 1101 0001 0101 0010 0100 1000 0110 1100 1(2) × 20 =

1.1110 1010 0100 1010 1111 1010 0010 1010 0100 1001 0000 1101 1001(2) × 2-1


7. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 0 (a positive number)

Exponent (unadjusted): -1

Mantissa (not normalized):
1.1110 1010 0100 1010 1111 1010 0010 1010 0100 1001 0000 1101 1001


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-1 + 2(11-1) - 1 =

(-1 + 1 023)(10) =

1 022(10)


9. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 1 022 ÷ 2 = 511 + 0;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1022(10) =

011 1111 1110(2)


11. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 52 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 1110 1010 0100 1010 1111 1010 0010 1010 0100 1001 0000 1101 1001 =

1110 1010 0100 1010 1111 1010 0010 1010 0100 1001 0000 1101 1001


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
011 1111 1110

Mantissa (52 bits) =
1110 1010 0100 1010 1111 1010 0010 1010 0100 1001 0000 1101 1001


Decimal number 0.957 603 280 698 573 651 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 1110 1010 0100 1010 1111 1010 0010 1010 0100 1001 0000 1101 1001

» Convert decimal 0.957 603 280 698 573 66 to 64 bit double precision IEEE 754 binary floating point representation standard

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How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders from the previous operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the list constructed above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100